6) The WWII era Calutron accelerated singly charged U 235 and 238 ions through 3

Question

6) The WWII era Calutron accelerated singly charged U 235 and 238 ions through 35kV into a 0.35T field where they went through 180 degree and were collected separately. Given: m235 = 235 x 1.7 x 10^-27kg m238 = 238 x 1.7 x 10^-27kg What is the radius of the orbit of each isotope in the Calutron and what is the separation of the two isotopes after traversing 180 degree. Hints a. Use the accelerating voltage to get kinetic energy as qV b. Use this to calculate the velocity c. Use masses, velocities ,charge and field to calculate r from mv^2/r = qvB d. Remember the separation of the isotopes is 2 x delta r.

Explanation / Answer

First qV = .5mv2

(1.6 X 10-19)(35000) = .5(235 X 1.7 X 10-27)v2

v = 1.67 X 105 m/s

r = mv/qB

r = (235)(1.7 X 10-27)(1.67 X 105)/(1.6 X 10-19)(.35)

r = 1.194 m for the U 235

(1.6 X 10-19)(35000) = .5(238 X 1.7 X 10-27)v2

v = 1.66 X 105 m/s

r = mv/qB

r = (235)(1.7 X 10-27)(1.66 X 105)/(1.6 X 10-19)(.35)

r = 1.187 m for the U 238

The separation after 180o of the two is (2)(1.194) - 2(1.187)

Separation after 180o = .0151 m (That is 1.51 cm)

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